| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Sample statistics from uniform |
| Difficulty | Standard +0.3 This is a straightforward application of uniform distribution properties and bias calculations. Part (a) requires showing E(X̄) ≠ α using the mean formula for uniform distributions, (b) is immediate subtraction, (c) involves simple algebra to make an unbiased estimator, and (d) uses the sample maximum as an estimator. All steps are standard textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{E(\bar{X}) = \mu\} = \frac{2\alpha+9+\alpha+3}{2}\) | M1 | Using the formula \(\left(\frac{b+a}{2}\right)\) or obtaining \(\frac{3\alpha+12}{2}\) or \(\frac{3\alpha}{2}+6\) |
| \(= \frac{3\alpha}{2}+6\) or \(\frac{3\alpha+12}{2} \neq \alpha\), so \(\bar{X}\) is a biased estimator | A1 | \(\frac{3\alpha}{2}+6\) or \(\frac{3\alpha+12}{2}\) and \(\neq \alpha\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{bias} = \frac{3\alpha}{2}+6-\alpha = \frac{1}{2}\alpha+6\) or \(\frac{\alpha+12}{2}\) (allow \(\pm\)) | B1ft | bias \(= \pm\left(\frac{1}{2}\alpha+6\right)\) or \(\pm\left(\frac{\alpha+12}{2}\right)\), or ft their \(\mu\) provided \(\mu \neq \alpha\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left\{E(Y) = \frac{2}{3}E(\bar{X})+k = \alpha \Rightarrow\right\} \frac{2}{3}\left(\frac{3\alpha}{2}+6\right)+k = \alpha\) | M1 | Sets \(\frac{2}{3}(\text{their } E(\bar{X}))+k = \alpha\). Mark can be implied |
| \(\{\alpha+4+k=\alpha \Rightarrow\}\ k=-4\) | A1 | \(k=-4\). Note \(k=-4\) with no working is M1 (implied) A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left\{\hat{\alpha} = \frac{2}{3}\bar{X}-4 \Rightarrow\right\}\ \hat{\alpha} = \frac{2}{3}(7.8)-4\ \{=1.2\}\) | M1 | An attempt to use the sample data to find \(\frac{2}{3}\bar{x}+\) "their \(k\)". Allow full expression for \(\bar{x}\) or \(\frac{\sum x}{n}\). Note from data \(\bar{x}=7.8\) |
| Max value \(= 2(1.2)+9\) | M1 | \(2\times\text{"their }\alpha\text{"}+9\) where \(\alpha\) is a function of the sample mean found by applying \(\frac{\sum x}{n}\) from the data |
| \(= 11.4\) | A1 | 11.4 cao. Also accept \(11\frac{2}{5}\) or \(\frac{57}{5}\) |
## Question 8:
*X follows a continuous uniform distribution over $[\alpha+3, 2\alpha+9]$; $Y = \frac{2\bar{X}}{3} + k$*
---
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{E(\bar{X}) = \mu\} = \frac{2\alpha+9+\alpha+3}{2}$ | M1 | Using the formula $\left(\frac{b+a}{2}\right)$ or obtaining $\frac{3\alpha+12}{2}$ or $\frac{3\alpha}{2}+6$ |
| $= \frac{3\alpha}{2}+6$ or $\frac{3\alpha+12}{2} \neq \alpha$, so $\bar{X}$ is a biased estimator | A1 | $\frac{3\alpha}{2}+6$ or $\frac{3\alpha+12}{2}$ **and** $\neq \alpha$ |
**[2]**
---
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{bias} = \frac{3\alpha}{2}+6-\alpha = \frac{1}{2}\alpha+6$ or $\frac{\alpha+12}{2}$ (allow $\pm$) | B1ft | bias $= \pm\left(\frac{1}{2}\alpha+6\right)$ or $\pm\left(\frac{\alpha+12}{2}\right)$, or ft their $\mu$ provided $\mu \neq \alpha$ |
**[1]**
---
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left\{E(Y) = \frac{2}{3}E(\bar{X})+k = \alpha \Rightarrow\right\} \frac{2}{3}\left(\frac{3\alpha}{2}+6\right)+k = \alpha$ | M1 | Sets $\frac{2}{3}(\text{their } E(\bar{X}))+k = \alpha$. Mark can be implied |
| $\{\alpha+4+k=\alpha \Rightarrow\}\ k=-4$ | A1 | $k=-4$. Note $k=-4$ with no working is M1 (implied) A1 |
**[2]**
---
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left\{\hat{\alpha} = \frac{2}{3}\bar{X}-4 \Rightarrow\right\}\ \hat{\alpha} = \frac{2}{3}(7.8)-4\ \{=1.2\}$ | M1 | An attempt to use the sample data to find $\frac{2}{3}\bar{x}+$ "their $k$". Allow full expression for $\bar{x}$ or $\frac{\sum x}{n}$. Note from data $\bar{x}=7.8$ |
| Max value $= 2(1.2)+9$ | M1 | $2\times\text{"their }\alpha\text{"}+9$ where $\alpha$ is a function of the sample mean found by applying $\frac{\sum x}{n}$ from the data |
| $= 11.4$ | A1 | 11.4 cao. Also accept $11\frac{2}{5}$ or $\frac{57}{5}$ |
**Note:** $2(10.6)+9 = 30.2$ is M0M0A0
**[3]**
**Total: 8 marks**8. The random variable $X$ has a continuous uniform distribution over the interval $[ \alpha + 3,2 \alpha + 9 ]$ where $\alpha$ is a constant.
The mean of a random sample of size $n$, taken from this distribution, is denoted by $\bar { X }$
\begin{enumerate}[label=(\alph*)]
\item Show that $\bar { X }$ is a biased estimator of $\alpha$
\item Hence find the bias, in terms of $\alpha$, when $\bar { X }$ is used as an estimator of $\alpha$
Given that $Y = \frac { 2 \bar { X } } { 3 } + k$ is an unbiased estimator of $\alpha$
\item find the value of the constant $k$
A random sample of 8 values of $X$ is taken and the results are as follows\\
4.8\\
5.8\\
6.5\\
7.1\\
8.2\\
9.5\\
9.9\\
10.6
\item Use the sample to estimate the maximum value that $X$ can take.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{3b6aaa8a-aeac-4a44-820b-e82f317d0c85-28_2633_1828_119_121}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2017 Q8 [8]}}